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# Integration problem, completely stuck Watch

1. I've been stuck on this question for the last 40 minutes and while it seems fairly straightforward I just can't get anywhere with it and so thought I would ask the wonderful people of TSR for some help!

Also, please excuse me typing it out, I never managed to figure out how to use LATEX.

I need to find the integral from 0 to infinity of exp(-y)cos(xy)dy

As far as i'm aware using integration by parts is a must here so I took the following:

u = exp(-y)
v' = cos(xy)
SO
u' = -exp(-y)
v = [sin(xy)]/x

which gives (using I to stand for the integral from 0 to infinity)

[exp(-y)sin(xy)]/x - I[-exp(-y)sin(xy)]/x

From then I need to use parts again and I take:

u = sin(xy)/x
v' = -exp(-y)/x
SO
u' = cos(xy)
v = exp(-y)/x

Which then gives me:

[exp(-y)sin(xy)]/x - [[exp(-y)sin(xy)]/x^2 - I[cos(xy)exp(-y)/x]]

But from here it all seems to fall apart.

I know what the answer needs to be, I know it has to be:

exp(-y)[xsin(xy)-cos(xy)]
------------------------------
1+x^2

but I have no idea how to get to there.

Thanks very much.
2. (Original post by mackemforever)
I never managed to figure out how to use LATEX.
http://www.thestudentroom.co.uk/wiki/LaTex

Give it five minutes.

I expect you already know, but after doing integration by parts, twice, the original integral will reappear.

Call it I and you get

I=stuff -x^2 I
3. (Original post by mackemforever)
I need to find the integral from 0 to infinity of exp(-y)cos(xy)dy
seems quite impossible (without doing something fancy that i've not learned yet)
4. When you do by parts the second time, you will end up with your original integral. If you say your initial integral = I then you get:

I = whatever - I

therefore 2I = whatever

therefore I = whatever/2
5. (Original post by ak9779)
When you do by parts the second time, you will end up with your original integral. If you say your initial integral = I then you get:

I = whatever - I

therefore 2I = whatever

therefore I = whatever/2
The problem being that I kept on ending up with I = whatever + I, therefore the left side vanishes.

That was why I was asking for somebody to run through it.
6. feynman integration
7. i can't understand what the question is, does the exp mean ??
8. (Original post by mph10)
i can't understand what the question is, does the exp mean ??
no, it means the exponential of -y, e.g. e^(-y) if that makes more sense.
9. (Original post by mackemforever)
no, it means the exponential of -y, e.g. e^(-y) if that makes more sense.
so the original post is

or is it

?
10. (Original post by DeanK22)
feynman integration
Well, that's something I've never heard of lol....
11. (Original post by mph10)
so the original post is

or is it

?
12. Are you integrating w.r.t. y?

EDIT: I've got the answer you've given above.

I strongly advise checking your signs in the steps in the middle.
13. (Original post by mackemforever)
are x and y independent variables?
14. i got to the same step as you, i think it just keeps on repeating integrals, so there must be another method, which unfortunately i cant remember
15. (Original post by mackemforever)
I've been stuck on this question for the last 40 minutes and while it seems fairly straightforward I just can't get anywhere with it and so thought I would ask the wonderful people of TSR for some help!

Also, please excuse me typing it out, I never managed to figure out how to use LATEX.

I need to find the integral from 0 to infinity of exp(-y)cos(xy)dy

As far as i'm aware using integration by parts is a must here so I took the following:

u = exp(-y)
v' = cos(xy)
SO
u' = -exp(-y)
v = [sin(xy)]/x

which gives (using I to stand for the integral from 0 to infinity)

[exp(-y)sin(xy)]/x - I[-exp(-y)sin(xy)]/x

From then I need to use parts again and I take:

u = sin(xy)/x
v' = -exp(-y)/x
SO
u' = cos(xy)
v = exp(-y)/x

Which then gives me:

[exp(-y)sin(xy)]/x - [[exp(-y)sin(xy)]/x^2 - I[cos(xy)exp(-y)/x]]

But from here it all seems to fall apart.

I know what the answer needs to be, I know it has to be:

exp(-y)[xsin(xy)-cos(xy)]
------------------------------
1+x^2

but I have no idea how to get to there.

Thanks very much.
Is this part of a double integral? If so we can treat x as if though it's a constant.
You should have let u be the exponential function on both occasions when you carried out the integration by parts. By swapping over for the second integration by parts, you are essentially reverting to what you started with. Also after dealing with the limits, your results should not involve y so your above answer that you seem to have been 'given' is completely wrong. If you do as I've said above, you should get to:
.
Which is almost what you got but not quite. If you let:

And rearrange the result I've written above for I, you have successfully carried out the integration (as soon as you evaluate the limits that result)
16. (Original post by marcusmerehay)
I strongly advise checking your signs in the steps in the middle.
This too, the signs in this probably get very confusing. Particularly when you're without paper...
17. (Original post by mackemforever)
I
u = exp(-y)
v' = cos(xy)
SO
u' = -exp(-y)
v = [sin(xy)]/x
Something they really should teach in A level:

http://en.wikipedia.org/wiki/Integration_by_parts#Liate_rule

the exponential is always v' so just switch your u and v' assignment and it should solve your infinite recursion problem.
18. above post by Farhan.Hanif93 is correct^^^^
19. (Original post by whiplash)
Something they really should teach in A level:

http://en.wikipedia.org/wiki/Integra...rts#Liate_rule

the exponential is always v' so just switch your u and v' assignment and it should solve your infinite recursion problem.

that does not work in this situation, try it, you will see.
20. (Original post by whiplash)
Something they really should teach in A level:

http://en.wikipedia.org/wiki/Integra...rts#Liate_rule

the exponential is always v' so just switch your u and v' assignment and it should solve your infinite recursion problem.
That's not true and I've had this discussion with someone else on here before. It doesn't matter whether you choose the trig function or the exp function as u, as long as you choose the same type of function for u when you integrate by parts for a second time.

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